Let $(X,A,\mu)$ be a positive measure space and let $f_n(x)\to_{a.e} 0$ and $\lim_{n \to \infty}\int_{x}f_nd\mu=0$ Then there must be integrable $g$ s.t $f_n\leq g$ for all $n$.
Counterexample: in $(\mathbb{R},S,m)$ we will define $$ f_n(x)= \begin{cases} \frac{1}{n}, n-1\leq x\leq n\\ 0, \text{otherwise}\\ \end{cases} $$
$f_n\to_{a.e}0$ as for all $x_o\in[n-1,n]$ $f_n(x_0)=0$ countable many times (?)
and $\int_{\mathbb{R}}f_nd\mu=\int_{\mathbb{R}}\frac{1}{n}1_{[n-1, n]}d\mu=\frac{1}{n}\to 0$ (? as $f_n>0$ so we can write it as a linear combination of simple function times the director function?)
Now assume that $g$ is measurable and for all $n$ we have $f_n\leq g$ then:
$\int_{\mathbb{R}}f_ndm\leq \int_{\mathbb{R}}gdm$ due to the monotonic property of the integral
Therefore:
$$\int_{\mathbb{R}}|g|dm\geq_{(1)} \int_{\mathbb{R}}gdm = _{(2)}\sum_{n=1}^{\infty}\int _{\mathbb{R}}g1_{[n-1,n)}dm\geq \sum_{n=1}^{\infty}\int _{\mathbb{R}}f_n1_{[n-1,n)}dm= \sum_{n=1}^{\infty}\frac{1}{n}=\infty$$
Where are moves $1,2$ are correct? and do the claims where there are $?$ are correct?